Dual of product isometric to product of dual Banach spaces

Question

I'm tying to show that if $X$ is a real Banach space then there is an isometric injection $$\tau: X' \times X' \to (X \times X)' $$

Where $X \times X$ has the norm $\|(x_1, x_2)\| = \|x_1\| + \|x_2\|$ and $X' \times X'$ has norm $\|(\phi_1, \phi_2)\| = \max\{\|\phi_1\|, \|\phi_2\|\}$

I've defined $\tau$ by $\tau(\phi_1, \phi_2)(x_1,x_2) = \phi_1(x_1) + \phi_2(x_2)$ and I'm now trying to show this is a well defined isometry.

I can easily show that $\tau$ is well defined and linear and by direct calculation I have that $\|\tau(\phi_1, \phi_2)\| \le \|(\phi_1, \phi_2)\|$.

Now I'm having trouble showing the other direction of this inequality to ensure $\tau$ is an isometry. I'm trying to make a smart choice for $(x_1, x_2)$ to show this but I'm not sure what values to pick!

Thanks

Answer 1

You have $\lVert (\phi_1, \phi_2) \rVert = \max\{ \lVert \phi_1 \rVert, \lVert \phi_2 \rVert \} = \lVert \phi_i \rVert$ for some $i \in \{ 1, 2 \}$. Assume for simplicity that $i = 1$.

Then your choice of $(x_1, x_2)$ is $(x, 0)$, where $\phi_1$ almost attains its norm on $x$, i.e. $\lvert \phi_1(x) \rvert \ge (1 - \varepsilon) \, \lVert \phi_1 \rVert \,\lVert x \rVert$ for small $\varepsilon > 0$. This yields $$ \lVert\tau(\phi_1, \phi_2)\rVert \ge \frac{\lvert \phi_1(x_1) + \phi_2(x_2) \rvert}{\lVert (x_1, x_2) \rVert} = \frac{\lvert \phi_1(x)\rvert}{\lVert x \rVert} \ge (1-\varepsilon) \, \lVert \phi_1 \rVert = (1-\varepsilon)\, \lVert (\phi_1, \phi_2) \rVert $$